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Full text of "Treatise On Analysis Vol-Ii"

426 XV NORMED ALGEBRAS AND SPECTRAL THEORY closed, show that 7*1 is closed; if lm(T) is closed in F, then so is Im(ri); if 7*is an index operator, then so is 7\, and /(7\) = i(T) + dim(M). (By induction on dim(M).) 4. Let E, F, be two Hilbert spaces and let T be an unbounded operator from E to F. Suppose that T is closed and dom(r) dense in E. Let B be a continuous operator from EtoF. (a) Show that, for each x e Ker(T-h £), Deduce that, if Ker(T) has finite dimension and y(r)>0 and ||J?|| < y(T), then dim(Ker(r+B))^dim(Ker(D) (use Problem 9 of Section 6.3). Furthermore, Im(TH- B) is closed in F (consider the restriction of T-\- B to the orthogonal supple- ment of Ker(T) + Ker(T + B) in E). (b) Suppose that T is an index operator and that \\B\\ < y(T). Show that codim(Im(r-|-j&))^codim(Im(r)), and that i(T + B) ** i(T). (To prove the first inequality, consider (7*4- B)*. For the equality of the indices, reduce to the case where J"is injective, by considering the orthogonal supplement of Ker(T) and using Problem 3(c). Then observe that y(T+ XB) is a continuous function of A for 0 <; A ^ 1 .) 5. With the same general hypotheses as in Problem 4, suppose that B is a compact operator from E to F. (a) Show that, if Ker(r) is finite-dimensional and Im(T) is closed, then Ker(T+ B) is finite-dimensional and Im(T+ B) is closed. (Consider the orthogonal supplement M of Ker(T), and show first that M n Ker(r+ B) <= M n dom(jT) is finite-dimen- sional, by noting that there exists w>0 such that \\T' x\\ ^m\\x\\ for all x e M n dom(7), and using (5.9.4). Then take the orthogonal supplement N of M n Ker(r+ B) in M, and prove that there exists no sequence (#„) in N n dom(T) such that \\xj(\ = 1 for all /z and T(xn) + B(xn) tends to 0.) (b) Deduce from (a) that if T is an index operator then so is T+ 5, and that i(T+ B) — i(T), (To prove that Im(!T+ B) has finite codimension, consider T* -f- B*; then observe that At— *-/(T+ XB) is a finite continuous function of A on [0, 1] by virtue of Problem 4.) (c) Suppose that E = F and that there exists a regular value f o for T such that (jT— Cob"""1 is a compact operator. Show that, for all £ e C, the operator T—t,I has index 0; the spectrum of T is a denumerable discrete subset of C, all of whose points are eigenvalues of T, and (3"— &D"1 is a compact operator for all regular values ^ of T. (Note that and use the theory of Riesz (11.4.1).) Hence give an example of an operator of index 0 whose spectrum is empty (use Problem 9 of Section 11.6). With the notation of the proof of (15.12.8), prove that the spectrum of the operator Nn is contained in the annulus Sn : (n — l)i/2 2 l£l =» «1/2* For each pair of points (xn , yn) in Fn , there corresponds to Nn a measure m££ yn with support contained in Sn . If x =2 x* anc* y ^S yn are two points of E, with xn e Fn and yn e F^ , then thee)).